Question

When an object is placed $$20\ cm$$ from a concave mirror, a real image magnified $$3$$ times is formed. Find:
Where the object must be placed to give a virtual images three times the height of the object?

Answer 1

$$\large\textbf{Step 1: Calculate the focal length}$$

Given,

Magnification, m = -3 (Image is real)
Object distance$$,u$$$$=$$ $$-20cm$$ ; By sign convention as object is placed at the left side of the mirror.
we know,

Magnification, $$m=\frac{-v}{u}$$
or $$-3=\frac{-v}{-20}$$

Or $$v=-60cm$$

Thus the image is located at a distance of 60cm in front of the mirror.

Now using the mirror formula,

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$
$$\frac{1}{f}=\frac{1}{-60}+\frac{1}{-20}$$

$$f = -15 cm$$

$$\large\textbf{Step 2: Apply magnification formula}$$

Given,

Magnification, m = +3 (Image is virtual)

Magnification, $$m=\frac{-v}{u}=\frac{{{h}_{i}}}{{{h}_{o}}}$$
or $$+3=\frac{-v}{u}$$

Or $$v=-3u$$

Now using the mirror formula,
$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$$

Or $$\frac{1}{-15}=\frac{1}{-3u}+\frac{1}{u}$$
on solving $$u = -10cm$$

Hence, The Object located at a distance of 10 cm in front of the mirror.

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